1
HA 2 m P 1 = 9t A VA 4 P 1 = 9t RHA PA sin α PA PA cos α PA sin α PA PA cos α RHA P 1 P 9 to Bidang Li Bidang M Bidang N Reaksi Perle r Tinjau batang Bidang Mome Daerah A-C MX 1 = RA . X X 1 = 0 MA X 1 = 2.5 M Daerah B-C MX 2 = RB . X X 2 = 0 MA X 2 = 2.5 M Bidang Lintan Daerah A-C DX 1 = RA X 1 = 0 DA X 1 = 2.5 DC Daerah B-C DX 2 = RB X 2 = 0 DA X 2 = 2.5 DC Bidang Norm Daerah A-C NX 1 = - HA X 1 = 0 NA X 1 = 2.5 NC 2 m PB sin α PB PB cos α t B VB 3 t PB sin α PB PB cos α P 2 on intang Momen Normal etakan Soal 1 g yang miring Kontrol : 3,6+3,6-7,2 = 0 7,2-7,2 = 0 0 = 0 = 5,4 ton en X A = 3,6 . 0 = 0 ton MC = 3,6 . 2,5 = 9 ton X A = 3,6 . 0 = 0 ton MC = 3,6 . 2,5 = 9 ton ng A = 3,6 ton C = 3,6 ton = 3,6 ton C = 3,6 ton mal A = -5,4 ton C = -5,4 ton

Sta Tika

Embed Size (px)

DESCRIPTION

Teknik sipil

Citation preview

Page 1: Sta Tika

HA

2 m 2 m

P1 = 9t

A

VA

4

P1 = 9t

RHA

PA sin α PA PA cos α

PA sin α PA PA cos α

RHA

P1

P

9 ton

Bidang Lintang

Bidang Momen

Bidang Normal

Reaksi Perletakan

r

Tinjau batang yang miring

Bidang MomenDaerah A-C MX1 = RA . XX1 = 0 � MA = 3,6 . 0 = 0 tonX1 = 2.5 � MC = 3,6 . 2,5 = 9 tonDaerah B-C MX2 = RB . XX2 = 0 � MA = 3,6 . 0 = 0 ton X2 = 2.5 � MC = 3,6 . 2,5 = 9 ton Bidang LintangDaerah A-C DX1 = RA X1 = 0 � DA = 3,6 tonX1 = 2.5 � DC = 3,6 tonDaerah B-C DX2 = RB X2 = 0 � DA = 3,6 tonX2 = 2.5 � DC = 3,6 ton Bidang NormalDaerah A-C NX1 = - HA X1 = 0 � NA = X1 = 2.5 � NC =

2 m 2 m

PB sin α PB PB cos α

= 9t

B

VB

3

= 9t

PB sin α PB PB cos α

P2

9 ton

Bidang Lintang

Momen

Normal

Reaksi Perletakan Soal 1

Tinjau batang yang miring Kontrol :

3,6+3,6−7,2 = 0

7,2−7,2 = 0 0 = 0

= 5,4 ton

Bidang Momen

= RA . X MA = 3,6 . 0 = 0 ton

MC = 3,6 . 2,5 = 9 ton

= RB . X MA = 3,6 . 0 = 0 ton

MC = 3,6 . 2,5 = 9 ton

Bidang Lintang

DA = 3,6 ton DC = 3,6 ton

A = 3,6 ton C = 3,6 ton

Bidang Normal

A = -5,4 ton C = -5,4 ton